f(x)=sin²x+2sinxcosx+3cosx²x
=1+2cos²x+sin2x
=cos2x+2+sin2x
=√2sin(2x+π/4)+2
最大值为 √2+2
这时 2x+π/4=2kπ+π/2
x=kπ+π/8 k∈z
x的集合为 {x|x=kπ+π/8 k∈z}
单增:
2x+π/4∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-3π/8,kπ+π/8]
所以
单调增区间为
[kπ-3π/8,kπ+π/8] k∈z
f(x)=1+sin2x+ 2(cosx)^2=2+sin2x+cos2x=2+√2sin(2x+π/4)
1)当2x+π/4=2kπ+π/2, 即x=kπ+π/8时,f(x)取最大值2+√2
2)单调增区间:
2kπ-π/4<2x+π/4<2kπ+π/2, 即: kπ-π/4
2kπ+π/2<2x+π/4<2kπ+3π/2, 即: kπ+π/8
这里k都为任意整数。